A bacterial culture starts with 20,000 bacteria, and the number doubles every 40 min. Find the number T of minutes required for the culture to have 50,000 bacteria?

This is going to be an exponential function. Think about how [itex]2^x[/itex] changes when x changes (mentally substitute x=1, x=2, x=3). What if we had a number multiplied by [itex]2^x[/itex]? Think about it, and if you can't figure out where to go, post what you've tried to solve this.

parwana, this thread belongs in the homework help section. And you will receive help on a problem only if you show what you've tried and where you're stuck.

The idea of this forum is to clarify your doubts, not to do your work for you.

That said - and keeping arildno's explanation in mind - understand where you went wrong in your attempt. There is no rule that says : f(x+t) = f(x) + f(t) OR, even worse, f(x+t) = f(x) + t. The latter is what you applied in your attempt.

Show us what you've done on the other problems too...and we'll help you with them.

himanshu, how do you say it's 1 ? I think not. The derivative (slope) of a constant function ?

Thanks everyone, it was the latter one zorodius. Here are some more questions!

Given that f(x)= 10, find the ratio (f(x+t)-f(x))/t

What you want to do here is to figure out what you can write instead of f(x+t) and f(x). For instance, you were given that f(x) = 10, so you can write the number 10 instead of f(x) in the problem. You can also figure out that f(x+t) is 10 as well, because f(x) doesn't change no matter what you put between the parentheses!

parwana said:

Find the roots of 4/h = (3/h^2)+1

I'll start this off:

[tex]\frac {4}{h} = \frac {3}{h^2} + 1[/tex]

[tex]0 = \frac {3}{h^2} + 1 - \frac {4}{h}[/tex]

[tex]0 = 3 + h^2 - 4h = h^2 - 4h + 3[/tex]

Can you see where to go from there?

parwana said:

One pipe can fill a swimming pool in 5 hours less than another. Together they fill the swimming pool in 5 hours. How long would it take each pipe to fill the tank alone?

First, let's give names to the quantities we're talking about.

A is the rate at which the fast pipe fills up pools. (in pools / hour)
B is the rate at which the slow pipe fills up pools. (in pools / hour)
Ta is the time it would take A to fill up a whole pool. (in hours)
Tb is the time it would take B to fill up a whole pool. (in hours)

Then "the rate at which a pipe fills up pools" times "the number of hours it takes that pipe to fill up a pool" should be equal to one pool, and so:

[tex]1 = A \cdot T_a[/tex]
[tex]1 = B \cdot T_b[/tex]

We also know, from the problem, that the time for the fast pipe to fill up a pool is five hours less than the time for the slow pipe to fill up a pool, so:

[tex]T_a = T_b - 5[/tex]

And we know that, working together, they fill one pool in five hours. Working together, their total rate is the sum of their individual rates, and:

[tex]1 = (A + B) 5[/tex]

So A + B is the rate of both pipes working together, and 5 is the number of hours it takes them to fill one pool.

Can you see where to go from there?

parwana said:

Given that f(x)= cos(x+1) and g(x) = ((x^2)-1)/2, find (f o g)(x)

I'll show an example of how a similar problem could be solved:

Given that a(x) = 2x and b(x) = x + 1, find (a o b)(x).

We replace (a o b)(x) with the equivalent a(b(x)). For the innermost function, we know that b(x) = x + 1, so a(b(x)) = a(x+1).

a(x) = 2x, meaning that to evaluate function a, we multiply whatever is between the parentheses by two. We had a(x+1), so a(x+1) = 2 * (x+1) = 2x + 2.

Be particularly aware of that when f(x+t) and f(x) are used in the ratio, x,x+t is meant to be arbitrary numbers, and not as a "variable" used in the definition of f.
So, (10-10)/t=0, for all (non-zero) t, since 10-10=0